# Find instantaneous velocity

1. Nov 21, 2006

### Fusilli_Jerry89

A particle moves in a line so its position at any time in t ≥0 is given by the function s(t)=t²-6t+5,
where s is measured in meres and t is measured in seconds.

a) find the displacement during the first 6 seconds.
All i did was input 0 into the equation solve, then do the same with 6 and subtract the two. I got 0. Is this the right way to do that, os is that too "physics like"?

b)Find avg. velocity during first 6 seconds.
i basically did the same thing with (y2-y1)/(x2-x1) and got 0.

c) Find instantaneous velocity when t=4.
the derivative of the function is 2t-6 so:
s(4)=2(4)-6
=2

d) Find the acceleration of the particle when t=4
the derivative of 2t-6 is 2 so is it 2?

e) At what values does the particle change direction?
I got when t=3 but only by looking at the graph. How do you do this by calculus?

f) Where is the particle when s is a minimum?
again by using the graph

2. Nov 21, 2006

### Staff: Mentor

I moved this thread to the homework forums. Jerry, you've posted in the homework forums before, so you should know where to post. No warning points issued.

You also know you need to show your work. Please show your initial work on a-f and we'll try to help.

EDIT -- Oops, sorry, I missed the part where he showed some work on each...

Last edited: Nov 22, 2006
3. Nov 22, 2006

### Mindscrape

Looking at a graph that you've made is perfectly valid. Regardless, if you think about it, finding when the object changes direction will be at the max or minimum of the displacement function. To find a maximum or minimum you take the derivative and set it equal to zero, then solve for t.

For part e) you should probably say the value of time, and then the corresponding displacement, velocity, and acceleration at that time. Part f) is just making sure you realize you found a min in e).

4. Nov 22, 2006

### Staff: Mentor

for -a-, use 6s in the equation...

5. Nov 22, 2006

### HallsofIvy

Staff Emeritus
As long as you get the correct answer! No, evaluating a function is not too "physics like". Even mathematicians are allowed to do arithmetic!

Yes, the particle moved out to a distance -4 meters at t= 3 seconds and then came back. Since it was right back where it started, its average velocity was 0.

Good. The derivative is the instantaneous velocity here. Actually, if I were you teacher, I would deduct a point. The correct answer is " 2 meters per second". Do you see the difference?

Well, more precisely, the acceleration, when t= 4 is 2 meters per second per second.

You can't throw a car into reverse when you are doing 80 mph! The only way a particle can "turn around" is if its velocity changes sign and that means the velocity must be 0 at that point. You have already determined that the velocity at time t is 2t- 6. For what t is 2t- 6= 0?

If, at any t, the velocity is positive, the particle is still moving to the right and just a moment before, s was lower: not a minimum. If at any t, the velocity is negative, the particle is moving to the left and a moment later s will be lower: not a minimum. The velocity must be 0 at s minimum. You've already determined what t that is. What is s for that t?